When a 1984 Alfa Romeo Spider sports car accelerates at the maxim...

Question

When a 1984 Alfa Romeo Spider sports car accelerates at the maximum possible rate, its motion during the first 20 s is extremely well modeled by the simple equationwhere P = 3.6 ✕ 10⁴ watts is the car's power output, m = 1200 kg is its mass, and v𝓍 is in m/s. That is, the square of the car's velocity increases linearly with time.a. Find an algebraic expression in terms of P, m, and t for the car's acceleration at time t.

Accepted Answer

Hey, everyone. In this problem, the equation V X squared equals N multiplied by P no divided by M all multiplied by T describes the velocity of an engine powered device for time T one equals zero and T two equals 30 seconds. During this time interval, the engine is at its peak acceleration. The equation contains an engine's power and multiples of P. OK. So we have N multiplied by peanut in watts. The device is mass M in kilograms and B in meters per second. The equation is interpreted as V squared having a direct relationship with T. We're asked to derive an expression for the acceleration in terms of P O M and T that can be used to determine acceleration at a time T less than or equal to 30 seconds. We're given four answer choices. Option A A X is equal to the square root of N P knot divided by two M T. Option B A X is equal to the square root of N P knot divided by four M T. Option C A X is equal to the square root of N P not T divided by four M. And option D A X is equal to the square root of N P knot T divided by two M T. Now we're given an expression or an equation for the velocity V X squared, we're looking for the acceleration. So let's recall how the acceleration and the velocity are related. The acceleration A X is going to be equal to the derivative of the velocity with respect to time. So we have A X is equal to D V X divided by D T. So what we need to do is take the derivative of our velocity equation. Now we were given V X squared not V X itself. So we need to take that equation and write it just as an equation for V X. And we have V X squared is equal to N P divided by M multiplied by T. And so V X, if we take the square root is going to be equal to the square root of N P not divided by M multiplied by T, now we can find our acceleration to the derivative. So we're gonna take the derivative. So A X is gonna be the derivative D divided by D T of this expression, the square root of N P knot divided by M multiplied by T. Now, when we have a square root, we can also write this as an exponent one half. I'm gonna do that. I think it's just a little bit cleaner and neater to write when we're doing our derivative. Um But if you like the square root better, you can write it like that as well. So we have peanut divided by M multiplied by T all to the exponent one half. Let me put brackets around this again. So it's clear that the one half is inside of that derivative. Now N P not divided by M to the exponent one half is just a constant. OK. So we have this constant multiplied by something else. When we're taking the derivative, we can pull that constant out in front. OK. And then worry about the derivative of T. So we have N P divided by M all to the exponent one half multiplied by the derivative D divided by D T of T to the exponent one half. Now when we're taking the derivative function of the exponent, what we're gonna do is we're gonna take the exponent, bring it down out in front and multiply and then reduce the exponent by one. So we get our constant stays the same and P nought divided by M to the exponent one half multiplied. The exponent comes out in front and multiplies one half T to the exponent. And now we're gonna subtract one from the exponent one half minus one gives us negative a half. So we get one half T to the exponent negative one half. All right, let's simplify, we have N P knot divided by M to the exponent one half. We have T to the exponent negative one half. OK? How do we deal with negative exponents? Well, what we can do is we can put the base and the exponent in the denominator and write it as a positive exponent. So this is equal to one half multiplied by one divided by T to the exponent one half. Now this one divided by T to the exponent one half, we can write this as one divided by T all to the exponent one half. OK. We have an exponent of one half in this constant term. So it'd be nice if we could write one divided by two and it is half con coefficient um in terms of an exponent of a half as well, so that we can simplify everything into one expression. Well, it turns out we can, OK, we can write this as N P not divided by M to the exponent one half multiplied by one divided by four T all to the exponent one half. OK? Because four to the exponent, one half gives us two, OK. Square root of four is two. We know that now we have two things. They both have an exponent of one half, they're being multiplied together. So we can combine these into a single term and we get N peanut divided by M times four T in the denominator all to the exponent one half. I remember at the beginning we switched our square root into an exponent of one half to make it a little bit easier when we were working with the derivative. Now we're gonna go back to the square root. OK. We have an exponent of one half. So this is equal to the square root of N P knot divided by, we're just gonna rearrange, we have M four T, we're just gonna write this as four M T. OK. Those things are multiplied so we can rearrange and do that multiplication in whichever order we'd like. All right. And this is the expression for the acceleration that we were asked to find. OK. We found that it's equal to the square root of N P knot divided by four M T. And if we look at our answer choices and compare what we found, we see that this corresponds with answer choice. B thanks everyone for watching. I hope this video helped see you in the next one.

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